2
$\begingroup$

A cursory search of the site showed no answer I was looking for. Essentially I have a few questions.

I know already that for $z_1, z_2 \in \mathbb{C}$, with arguments $\theta_1$ and $\theta_2$ respectively, we have $$ |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2|\cos(\theta_1 - \theta_2) $$

(for those who are interested, proof wiki has a simple proof here: https://proofwiki.org/wiki/Complex_Modulus_of_Sum_of_Complex_Numbers )

I have 2 questions. I think the first one is solved by me, but would like confirmation. The second question is a bit beyond me at the moment xD

  1. How would one generalize this for $z_1,...,z_k \in \mathbb{C},$ where we know the arguments of $z_i$: $$ \left| \sum_{i=1}^k z_i \right|^2? $$
  2. How would one generalize this for $\alpha_1,...,\alpha_k,z_1,...,z_k \in \mathbb{C},$ where we know the arguments of $z_i$ but not of $\alpha_i$: $$ \left| \sum_{i=1}^k \alpha_iz_i \right|^2? $$

My attempt so far:

1. Suppose for $z_1, ..., z_k \in \mathbb{C}$ with arguments $\theta_1, ...,\theta_k$ respectively, we have

$$ \left| \sum_{i=1}^k z_i \right|^2 = |z_1 + ... + z_k|^2 \\ =(|z_1|]\cos(\theta_1) + ... + |z_k|\cos(\theta_k))^2 + (|z_1|\sin(\theta_1) + ... + |z_k|\sin(\theta_k))^2 \\ = |z_1|^2 \cos^2\theta_1 + |z_1||z_2|\cos\theta_1\cos\theta_2 + ... |z_1|z_k|\cos\theta_1\cos\theta_k + \\ |z_2|^2 \cos^2\theta_2 + |z_1||z_2|\cos\theta_1\cos\theta_2 + ... |z_2|z_k|\cos\theta_2\cos\theta_k + \\ ... + \\ |z_k|^2\cos^2\theta_k + |z_1||z_k|\cos\theta_1\cos\theta_k + ... + |z_{k-1}||z_k|\cos\theta_{k-1}\cos\theta_k + \\ |z_1|^2 \sin^2\theta_1 + |z_1||z_2|\sin\theta_1\sin\theta_2 + ... |z_1|z_k|\sin\theta_1\sin\theta_k + \\ |z_2|^2 \sin^2\theta_2 + |z_1||z_2|\sin\theta_1\sin\theta_2 + ... |z_2|z_k|\sin\theta_2\sin\theta_k + \\ ... + \\ |z_k|^2\sin^2\theta_k + |z_1||z_k|\sin\theta_1\sin\theta_k + ... + |z_{k-1}||z_k|\sin\theta_{k-1}\sin\theta_k \\ = |z_1|^2(\cos^2\theta_1 + \sin^2\theta_1) + |z_2|^2(\cos^2\theta_2 + \sin^2\theta_2) + ... + |z_k|^2(\cos^2\theta_k + \sin^2\theta_k) + \\ 2|z_1||z_2|(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) +2|z_1||z_3|(\cos\theta_1\cos\theta_3 + \sin\theta_1\sin\theta_3) + \\ ... + \\ 2|z_{k-1}||z_k|(\cos\theta_{k-1}\cos\theta_k + \sin\theta_{k-1}\sin\theta_k) $$

Now use the fact that $\cos^2\theta + sin^2\theta = 1$ and the fact that $\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 = \cos(\theta_1 - \theta_2)$:

$$ \left| \sum_{i=1}^k z_i \right|^2 = \sum_{i=1}^k |z_i|^2 + \sum_{i \neq j}|z_i||z_j|\cos(\theta_i - \theta_j) $$

For 2. I have no idea where to start. I am asking because I am working through this paper on Quantum Clustering and Gaussian Mixtures https://arxiv.org/abs/1612.09199 and they give a simple answer: $$ \left| \sum_{i=1}^k \alpha_iz_i \right|^2 = \sum_{i=1}^k \alpha_i|z_i| \sum_{j=1}^k \bar{\alpha_j}|z_j|\cos(\theta_i - \theta_j) $$ is this true? How would I show it?

For those who are interested, in the paper $|z_k|$ is the unnormalized Gaussian Distribution with parameters $(\mu_k, \mathbf{C}_k)$ given for some $d$-dimensional point $p$ by $$ |z_k| = \frac{1}{\sqrt{Z_k}}e^{-\frac{1}{4}(p - \mu_k)^T \mathbf{C}^{-1}_k(p-\mu_k)} \\ Z_k = (2\pi)^{\frac{d}{2}}|\mathbf{C}_k|^{\frac{1}{2}} $$ whereas $z_k$, the complex number, is given by $$ z_k = \frac{1}{\sqrt{Z_i}}e^{-\frac{1}{4}(p - \mu_k)^T \mathbf{C}^{-1}_k(p-\mu_k)}e^{-i\phi_k} $$ When I programmed this up I encountered some errors (i am getting negative probabilities), and now I'm trying to determine if its because of my code or if there's an error in the paper. I think it might because the exponent with the phase is $-i\phi_k$ instead of $i\phi_k$. I'm checking out this hunch but it would be super helpful to know that the above summation of square modulus is correct.

Edit history:

  1. Removed coefficient of 2 from last term.
  2. Using saulspatz idea: $$ \left|\sum_{i=1}^k \alpha_iz_i \right|^2 = \sum_{i=1}^k \alpha_i z_i \sum_{j=1}^k \overline{\alpha_j z_j} \\ = \sum_{i=1}^k \alpha_i |z_i|(\cos\theta_i + i\sin\theta_i) \sum_{j=1}^k \overline{\alpha_j} |z_j|(\cos\theta_j - i\sin\theta_j) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|(\cos\theta_i + i\sin\theta_i)(\cos\theta_j - i\sin\theta_j) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| (\cos\theta_i\cos\theta_j -i\cos\theta_i\sin\theta_j + i\cos\theta_j\sin\theta_i +\sin\theta_i\sin\theta_j) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| (\cos\theta_i\cos\theta_j + \sin\theta_i\sin\theta_j + i\cos\theta_j\sin\theta_i - i\cos\theta_i\sin\theta_j) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| (\cos(\theta_i - \theta_j) + i\sin(\theta_i - \theta_j)) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\cos(\theta_i - \theta_j) + \overline{\alpha_j} |z_j|i\sin(\theta_i - \theta_j) \\ = \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\cos(\theta_i - \theta_j) + i\sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\sin(\theta_i - \theta_j) $$ Presumably this imaginary part equals 0, but I'm really at a loss as to how to show that >.<

In any case, now I see that the real part must be as was shown in the paper, meaning I have a bug in my code... whoops.

$\endgroup$
  • $\begingroup$ It's the square of a modulus, so undoubtedly the imaginary parts should somehow cancel away... but it's hard to see how. $\endgroup$ – Marko Apr 4 '18 at 6:54
2
$\begingroup$

Start with $$\left|\sum_{i=1}^k{a_iz_i}\right|^2=\sum_{i=1}^k{a_i}{z_i}\sum_{j=1}^k\overline{{a_j}{z_j}} $$

Note that the coefficient of $2$ is incorrect in the formula you give. When you sum over $i\ne j$ you sum over $(2,3)$ and $(3,2)$ separately.

EDIT You're making things too hard. I'm going to change the index of summation from $i$ to $m$ to avoid confusion with the imaginary unit.

$$\left|\sum_{m=1}^k{a_mz_m}\right|^2= \sum_{m=1}^k{a_mz_m}\sum_{j=1}^k\overline{{a_jz_j}}= \sum_{m=1}^k{a_mz_m}\sum_{j=1}^k\overline{{a_jz_j}}=\\ \sum_{m=1}^k{a_m|z_m|e^{i\theta_m}}\sum_{j=1}^k{\overline{a_j}|z_j|e^{-i\theta_j}}= \sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|e^{i(\theta_m-\theta_j)}}=\\ \sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\cos(\theta_m-\theta_j)}+i\sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\sin(\theta_m-\theta_j)}=\\ \sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\cos(\theta_m-\theta_j)} $$ because $$\left|\sum_{m=1}^k{a_mz_m}\right|^2\text{ is real.}$$

So yes, the imaginary part is $0$, but it's usually easier to use $e^{i\theta}$ than $\cos\theta+i\sin\theta.$

$\endgroup$
  • $\begingroup$ Right that was a silly mistake from me, thanks. I'm busy looking at what you just wrote and seeing how much progress I can make on it. $\endgroup$ – Marko Apr 4 '18 at 4:11
  • 1
    $\begingroup$ @Marko The only part of math I was ever really good at was silly mistakes. $\endgroup$ – saulspatz Apr 4 '18 at 4:16
  • $\begingroup$ Hmmm either I'm stuck or the paper does indeed have a mistake... $\endgroup$ – Marko Apr 4 '18 at 6:36
  • $\begingroup$ @Marko I just looked at the obvious way to start. I'll check it out. $\endgroup$ – saulspatz Apr 4 '18 at 13:16
  • 1
    $\begingroup$ @Marko I think I've got it right now. Thanks for the heads up. $\endgroup$ – saulspatz Apr 6 '18 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.